FROM A REGULARLY STRATIFIED MEDIUM
In this we assume un= vn+1/vn, so that
497
l+1 + ?vn=0...........5,.......(31)
The solution of (31) is
where p + q = r2-t2-!, pq = t2, .....................(32)
and H, K are arbitrary constants. Accordingly
BVi'i+i i ZfV>7i+i -"• f-i r -fi- (/ ,__.
-J/ — ---i.________ -i___ ^QQ 1
n Hpn + Kqn '.........................^oa'
in which there is but one constant of integration effectively.
This constant may be determined from the case of n = l, for which tt1 = 7'9—1. By means of (32) we get (p + l)H+(q + l) K=0,
so that ^ — L^^£_^yLt_t/_.^J > ..................(34)
and
or since by (32) ?>a = (p
fa = __I-__ ? ...........................(35)
where A/(-) = 6, A/(-—Wrt......................(36)
V \gy V Vp + 1/
In order to find tym we may put n= 1 in (17); and by use of (29), with m substituted for n, we get
T
and on reduction with use of (35), (32),
»--*afa-"^-.........................<37>
By putting TO = 0, we see that the upper sign is to be taken. The expressions thus obtained are those of Stokes:
0m_____ym ____________ f^R\
The connexion between a, b and r, t is established by setting, m = 1. Thus
b — 6—1 a — or*1 ft6 — or1 b~1
In Stokes' problem, where r, t, (j>, ty represent intensities, a and b are real. If there is no absorption, r + t = 1, so that a — 1, 6 — 1 are vanishing quantities.
In this case
r____t 1
b^l~a-l~a-l + b-l' E. vi. 32not introducing other than integral values of m and n. If we make m = 1 in (17),